Show-that-a-N-10-143-1-9-is-composite-and-b-N-has-two-factors-each-of-which-is-a-series-of-a-G-P-

Question Number 25046 by Tinkutara last updated on 02/Dec/17

S h o w t h a t

(a) N = \frac{10^{143} - 1}{9} i s c o m p o s i t e, a n d

(b) N h a s t w o f a c t o r s e a c h o f w h i c h i s

a s e r i e s o f a G . P .

Answered by Rasheed.Sindhi last updated on 05/Dec/17

N (10) = \frac{10^{143} - 1}{9} = \frac{10^{143} - 1}{10 - 1}

In general

N (x) = \frac{x^{143} - 1}{x - 1}

= \frac{(x - 1) (x^{142} + x^{141} + x^{140} + \dots + x + 1)}{(x - 1)}

= x^{142} + x^{141} + x^{140} + \dots + x + 1

So,

N (10) = 10^{142} + 10^{141} + \dots + 10 + 1

[N (10) = 111 \dots 11 (143 times)

143 = 11 \times 13

So the number can be grouped either

11 groups of 13^{'} {ones}^{'} or 13 groups of 11^{'} {ones}^{'} .]

= (10^{12} + 10^{11} + \dots 10 + 1) . 10^{130}

+ (10^{12} + 10^{11} + \dots 10 + 1) . 10^{117}

+ (10^{12} + 10^{11} + \dots 10 + 1) . 10^{104}

+ \dots + (10^{12} + 10^{11} + \dots 10 + 1) . 10^{13}

+ (10^{12} + 10^{11} + \dots 10 + 1) . 1

= (10^{12} + 10^{11} + . . 10 + 1) (10^{130} + 10^{117} + \dots 10^{13} + 1)

(a) Hence the number is composite .

(b) Both factors are geometric series .

10^{12} + 10^{11} + 10^{10} \dots + 10 + 1 : common ratio 10^{- 1} .

10^{130} + 10^{117} + 10^{104} + \dots + 10^{13} + 1 : common ratio 10^{- 13}

Commented by Tinkutara last updated on 05/Dec/17

Thank you Sir!

T h e r e i s o n e m o r e a n s w e r :

N = (1 + 10^{11} + 10^{22} + \dots + 10^{132}) (1 + 10 +

10^{2} + \dots + 10^{10})